It is a simultaneous equation and when solved its solutions are x = 71/26 and y = 50/13
Another straight line equation is needed such that both simultaneous equations will intersect at one point.
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
They are simultaneous equations and their solutions are x = 41 and y = -58
No. It's a quadratic equation, and it has two solutions.
It is a simultaneous equation and when solved its solutions are x = 71/26 and y = 50/13
If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7
The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11
Another straight line equation is needed such that both simultaneous equations will intersect at one point.
Through a process of elimination and substitution the solutions are s = 8 and x = 5
There are 120 solutions.
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
There are no real solutions because the discriminant of the quadratic equation is less than zero.
They are simultaneous equations and their solutions are x = 41 and y = -58
It is a quadratic equation and its solutions can be found by using the quadratic equation formula.
No. It's a quadratic equation, and it has two solutions.
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